None of the first $k$ primes divide $p_1p_2\dots p_k +1$, but clearly $p_{k+1} \geq p_k + 2$ so there are no solutions? Checking $k=1$ for the $(2,3)$ case doesn't work either.

k = 4 (2 + 1)(3 + 1)(5 + 1)(7 + 1) = 24^2

Arriving at 13 we’ll have required a second factor of 7 and not further down it’s factor of 5 for prime 19, which will only return after the next prime ending in 9 which is 29 Second 7 will appear for p = –1 mod 14, and the next after 13 would be 41 , and between the primes 19 and 41, there’ll be even more other larger primes that would need their own pairing , like 19 from 37 that will only resurface for prime 113

There are no solution. Consider putting the one on the other side $$p_1p_2\dots p_k +1=x^n$$$$\Rightarrow p_1p_2\dots p_k=x^n - 1$$So now let $p_i$ be the smallest prime that does not divide $x-1$, then we claim that $p_j$ does not divide $n$ for $j < i$. We will procede by contradiction.Use LTE on the RHS, then we would get (the inequality is also valid for p=2, you can also consider it separatly) $$v_p(RHS) \geq v_p(x-1)+v_p(n)\geq 2$$Absurd, so now $p_i$ must divide RHS, but now consider the order of $x$ mod $p_i$, it then must divide $n$, but it cannot be 1, and so it should be a factor of $n$ less than $p_i$, but in the factorization it has no factor less than $p_i$ (by above).